Chapter 26 - Electric Field. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University

Transcription

1 Chapte 6 lectic Field A PowePoint Pesentation by Paul. Tippens, Pofesso of Physics Southen Polytechnic State Univesity 7

2 Objectives: Afte finishing this unit you should be able to: Define the electic field and explain what detemines its magnitude and diection. Wite and apply fomulas fo the electic field intensity at known distances fom point chages. Discuss electic field lines and the meaning of pemittivity of space. Wite and apply Gauss's law fo fields aound sufaces of known chage densities.

3 The Concept of a Field A field is defined as a popety of space in which a mateial object expeiences a foce. Pm. F Above eath, we say thee is a gavitational field at P. Because a mass m expeiences a downwad foce at that point. No foce, no field; No field, no foce! The diection of the field is detemined by the foce.

4 The Gavitational Field A F B F If g is known at evey point above the eath then the foce F on a given mass can be found. Conside Note that points the foce A and F is B eal, above, but the suface field is just of the a convenient eath just points way of in descibing space. space. The field at points A o B might be found fom: g F m The magnitude and diection of the field g is depends on the weight, which is the foce F.

5 The lectic Field 1. Now, conside point P a distance fom +Q.. An electic field exists at P if a test chage +q has a foce F at that point. 3. The diection of the is the same as the diection of a foce on + (pos) chage. 4. The magnitude of is given by the fomula: +q P + + Q F q +. F lectic Field N ; Units C

6 Field is Popety of Space F +q Q lectic Field Foce on +q is with field diection. Foce on q is against field diection. q F. + + Q lectic Field The field at a point exists whethe thee is a chage at that point o not. The diection of the field is away fom the +Q chage.

7 Field Nea a Negative Chage +q +. F Q lectic Field Foce on +q is with field diection. Foce on q is against field diection. F q. Q lectic Field Note that the field in the vicinity of a negative chage Q is towad the chage the diection that a +q test chage would move.

8 The Magnitude of Field The magnitude of the electic field intensity at a point in space is defined as the foce pe unit chage (N/C) that would be expeienced by any test chage placed at that point. lectic Field Intensity F N ; Units q C The diection of at a point is the same as the diection that a positive chage would move IF placed at that point.

9 xample 1. A + nc chage is placed at a distance fom a 8 C chage. If the chage expeiences a foce of 4 N, N what is the electic field intensity at point P? Fist, we note that the diection of is towad Q Q (down). F q 4 N 9 x 1 C +q + nc +. P Q 4 N 8 C lectic Field = x 1 1 N/C Downwad Note: The field would be the same fo any chage placed at point P.. It is a popety of that space.

10 xample. A constant field of 4, N/C is maintained between the two paallel plates. What ae the magnitude and diection of the foce on an electon that passes hoizontally between the plates.c The field is downwad, and the foce on e is up. F ; F q q F q e 19 4 F e. N C (1.6 x 1 C)(4 x 1 ) F = 6.4 x N, Upwad e

11 The Field at a distance fom a single chage Q Conside a test chage +q placed at P a distance fom Q. The outwad foce on +q is: The electic field is theefoe: F kqq Q q F kqq q kq q F +. P kq

12 xample 3. What is the electic field intensity at point P,, a distance of 3 m fom a negative chage of 88 nc? =?. 3 m Q 8 nc P Fist, find the magnitude: kq (9 x 1 )(8 x 1 C) 9 Nm 9 C = 8. N/C (3 m) The diection is the same as the foce on a positive chage if it wee placed at the point P: towad Q. = 8. N, towad Q

13 The Resultant lectic Field. The esultant field in the vicinity of a numbe of point chages is equal to the vecto sum of the fields due to each chage taken individually. Conside fo each chage. Vecto Sum: = q 1 R 1 q 3 3 A + q Magnitudes ae fom: kq Diections ae based on positive test chage.

14 xample 4. Find the esultant field at point A due to the 33 nc chage and the +6 nc chage aanged as shown. q 1 3 cm 3 nc 1 A 5 cm +6 nc 4 cm + q fo each q is shown with diection given. kq ; kq (9 x 1 )(3 x 1 C) 9 Nm 9 C 1 (3 m) (9 x 1 )(6 x 1 C) 9 Nm 9 C (4 m) Signs of of the chages ae used only to to find diection of of

15 xample 4. (Cont.)Find the esultant field at point A.. The magnitudes ae: q 1 3 cm 3 nc 1 A 5 cm +6 nc 4 cm + q (9 x 1 )(3 x 1 C) 9 Nm 9 C 1 (3 m) (9 x 1 )(6 x 1 C) 9 Nm 9 C (4 m) 1 = 3. N, West = 3.38 N, Noth Next, we find vecto esultant R R ; tan R 1 1 R 1

16 xample 4. (Cont.)Find the esultant field at point A using vecto mathematics. R 1 1 = 3. N, West = 3.38 N, Noth Find vecto esultant R (3. N) (3.38 N) 4.5 N; tan = 48.4 N of W; o = N 3. N Resultant Field: R = 4.5 N; 131.6

17 lectic Field Lines lectic Field Lines ae imaginay lines dawn in such a way that thei diection at any point is the same as the diection of the field at that point. + + Q Q Field lines go away fom positive chages and towad negative chages.

18 Rules fo Dawing Field Lines 1. The diection of the field line at any point is the same as motion of +q at that point.. The spacing of the lines must be such that they ae close togethe whee the field is stong and fa apat whee the field is weak. 1 + q 1 q R

19 xamples of Field Lines Two equal but opposite chages. Two identical chages (both +). Notice that lines leave + chages and ente chages. Also, is stongest whee field lines ae most dense.

20 The Density of Field Lines Gauss s Law: The field at at any point in in space is is popotional to to the line density at that point. Gaussian Suface Line density A N Radius N A

21 Line Density and Spacing Constant Conside the field nea a positive point chage q: Then, imagine a suface (adius ) suounding q. Radius is popotional to N/ N/A and is equal to kq/ at any point. N kq ; A Gaussian Suface N Whee is: A Define as spacing constant. Then: 1 4 k

22 Pemittivity of Fee Space The popotionality constant fo line density is known as the pemittivity and it is defined by: 1 C 4 k Nm x 1 Recalling the elationship with line density, we have: N o N A A Summing ove entie aea A gives the total lines as: N = o A

23 xample 5. Wite an equation fo finding the total numbe of lines N leaving a single positive chage q. Radius Gaussian Suface Daw spheical Gaussian suface: N A and N A Substitute fo and A fom: kq q ; A = 4 4 q N A (4 ) N = 4 o qa = q Total numbe of of lines is is equal to to the enclosed chage q. q.

24 Gauss s s Law Gauss s s Law: The net numbe of of electic field lines cossing any closed suface in in an outwad diection is is numeically equal to to the net total chage within that suface. N Aq If we epesent q as net enclosed q positive chage,, we can wite A ewite Gauss s s law as:

25 xample 6. How many electic field lines pass though the Gaussian suface dawn below. Fist we find the NT chage q enclosed by the suface: q q = (+8 4 1) = +3 C N Aq 4 C Gaussian suface +8 C q q 1 + q 4 1 C + q 3 +5 C N = +3 C = +3 x lines

26 xample 6. A solid sphee (R = 6 cm) having net chage +8 C is inside a hollow shell (R = 8 cm) having a net chage of 6 C.. What is the electic field at a distance of 1 cm fom the cente of the solid sphee? Daw Gaussian sphee at adius of 1 cm to find. N Aq q q = (+8 6) = + C q A qnet; A 8cm 1 cm Gaussian suface +8 C 6 C 6 cm 6 q x 1 C (4 ) (8.85 x 1 )(4 )(.1 m) 1 Nm C

27 xample 6 (Cont.) What is the electic field at a distance of 1 cm fom the cente of the solid sphee? Daw Gaussian sphee at Gaussian suface adius of 1 cm to find. N Aq q q = (+8 6) = + C q A qnet; A C (4 ) 1.5 x 1 8cm 1 cm 6 N C +8 C 6 C 6 cm = 1.5 MN/C

28 Chage on Suface of Conducto Since like chages epel, you would expect that all chage would move until they come to est. Then fom Gauss s s Law... Gaussian Suface just inside conducto Chaged Conducto Since chages ae at est, = inside conducto, thus: N Aq o = q All chage is is on suface; None inside Conducto

29 xample 7. Use Gauss s s law to find the field just outside the suface of a conducto. The suface chage density = q/a. Conside q inside the pillbox. lines though all aeas outwad. A q lines though sides cancel by symmety A Suface Chage Density The field is zeo inside the conducto, so = q o 1 A + o A = q A

30 xample 7 (Cont.) Find the field just outside the suface if = q/a = + C/m. Recall that side fields cancel and inside field is zeo, so that 1 q A A Suface Chage Density x 1 C/m 8.85 x Nm C = 6, N/C

31 Field Between Paallel Plates Q 1 + Q qual and opposite chages. Field 1 and to ight. Daw Gaussian pillboxes on each inside suface. Gauss s s Law fo eithe box gives same field ( 1 = ). A q q A

32 Line of Chage A 1 L A Field due to A 1 and A Cancel out due to symmety. A q q L A q q ; = L L q A ; A( ) L

33 xample 8: The lectic field at a distance of 1.5 m fom a line of chage is 5 x 1 4 N/C. What is the linea density of the line? L q L = 5 x 1 4 N/C = 1.5 m (8.85 x 1 )(1.5 m)(5 x 1 N/C) 1 C 4 Nm 4.17 C/m

34 Concentic Cylindes b b a Outside is like chaged long wie: a Gaussian suface 6 C b a a 1 1 cm b Fo a > b b Fo a b > > a

35 xample 9. Two concentic cylindes of adii 3 and 6 cm.. Inne linea chage density is +3 C/m and oute is 5 C/m.. Find at distance of 4 cm fom cente. Daw Gaussian suface 7 C/m + + between cylindes. a = cm b + + 3C/m (.4 m) b=6 cm +5 C/m = 1.38 x 1 6 N/C, Radially out

36 xample 8 (Cont.) Next, find at a distance of 7.5 cm fom cente (outside both cylindes.) Gaussian outside of both cylindes. a b ( 35) C/m (.75 m) 7 C/m a = 3 cm b=6 cm C/m = 5. x 1 5 N/C, Radially inwad

37 Summay of Fomulas The lectic Field F Intensity : : q kq N Units ae C The lectic Field kq Vecto Sum Nea seveal chages: Gauss s Law fo Chage distibutions. A q; q A

38 CONCLUSION: Chapte 4 The lectic Field